By an angle chase, we can see that $P$ moves on a circle tangent in $A,B$ to $CA,CB$ respectively. Let $Q$ be the second intersection of $CP$ with this circle. $PAQB$ is a harmonic cyclic quadrilateral, so, if $C'$ is the intersection between the tangent at $P$ to the circle and $AB$, we find $(PA,PB;PQ,PC')=-1$, meaning that $PQ$ is the symmedian of $PAB$, which is precisely what we want. I was in a hurry while writing this, but I'll clarify if needed.

Rewriting your problem with different notations: Let ABC be an isosceles triangle with AB = AC. Let P be a point inside the triangle ABC such that < PBC = < PCA. Let M be the midpoint of the segment BC. Prove that < BPM + < APC = 180°. Now, this problem was solved in posts #8 and #9 of http://www.mathlinks.ro/Forum/viewtopic.php?t=18258 (see the Lemma in post #8), with the only difference being that an additional assumption was used, namely the assumption that < PCB = < PBA; but this assumption is actually unnecessary, since it follows from the other assumptions of the problem (namely, since the triangle ABC is isosceles with AB = AC, we have < ACB = < ABC, and since < PBC = < PCA, we have < PCB = < ACB - < PCA = < ABC - < PBC = < PBA, so that < PCB = < PBA comes out as a consequence of the other assumptions of the problem). Darij

warut_suk wrote: Let a triangle $ABC$ satisfy $AC = BC$; in other words, let $ABC$ be an isosceles triangle with base $AB$. Let $P$ be a point inside the triangle $ABC$ such that $\angle PAB = \angle PBC$. Denote by $M$ the midpoint of the segment $AB$. Show that $\angle APM + \angle BPC = 180^{\circ}$. Take point Q such that APBQ is a parallellogram. Take point R on AQ such that ACB, APR are similar. (I assume R lies between A and Q, the other case is similar) Then BPRQ is cyclic, and ACP, ABR are similar. Hence BPQ = BRQ = 180 - ARB = 180 - APC. Hence result.

$S\in AB\cap CP$, $CA=CB=a$, $AB=b$, $m(\widehat {PCA})=u$, $m(\widehat {PCB})=v$, $m(\widehat {PAB})=m(\widehat {PBC})=x$, $m(\widehat {PAC})=m(\widehat {PBA})=y$. $\blacktriangle$ The theorem of Sinus in the triangle $APB\Longrightarrow \frac{PA}{PB}=\frac{\sin y}{\sin x}\ \ (1)$. $\blacktriangle$ The trigonometrical form of the Menelaus' theorem in the triangle $ABC$ for the inner point $P\Longrightarrow$ $\sin u\sin^2x=\sin v\sin^2y$ $\Longrightarrow$ $\frac{\sin u}{\sin v}=\left(\frac{\sin y}{\sin x}\right)^2\ \ (2)$. $\blacktriangle\ \frac{SA}{SB}=\frac{CA}{CB}\cdot \frac{\sin \widehat {ACS}}{\sin \widehat {BCS}}$ $\Longrightarrow$ $\frac{SB}{SC}=\frac{\sin u}{\sin v}\ \ (3).$ From the relations $(1)$, $(2)$, $(3)$ results: $\frac{SA}{SB}=\left(\frac{PA}{PB}\right)^2$, i.e. the ray $[PS$ is the symmedian from the vertex $P$ in the triangle $APB$, meaning $\widehat {APM}\equiv \widehat {BPS}\Longrightarrow$ $\boxed {\ m(\widehat {APM})+m(\widehat {BPC})=180^{\circ}\ }.$

The problem can be generalized as follow (which is well-known) Let $PAB$ be a triangle and $M$ be the midpoint of $AB$. Suppose $C$ is the intersection of the tangents of the circumcircle at $A$ and $B$. Then $\measuredangle{BPM}= \measuredangle{CPB}$. [Moderator edit: This is basically the assertion of http://www.mathlinks.ro/Forum/viewtopic.php?t=99571 .] Back to the problem , it is easy to see that $CA$ and $CB$ are the tangents of $(PAB)$ at $A$ and $B$.

See http://www.mathlinks.ro/Forum/viewtopic.php?t=55581 http://www.mathlinks.ro/Forum/viewtopic.php?t=43202 where are mentioned the properties of the harmonical quadrilateral.

Let us see the circumcircle of APB. Let v=<ABC. Use the center of the circumcircle as origin of coordinate and give the coordinates as: C (0,1) M (0, (cos v)^2) P (cos v*cos u, cos v*sin u) It is easy to obtain: PM/PC=cos v So, we can get: BC/PC=AM/PM. So, <BPC=<APM or <BPC+<APM=180. When <BPC=<APM, then <AMP=<BCP. So, <BPC=<BMC=90. As a result, <BPC+<APM=180 too.

P is on circle (U) tangent to BC at B and passing through A(then $ \angle PAB = \angle PBC$). $ \triangle ABC$ is isosceles $ \Longrightarrow$ (U) is also tangent to CA at A, AB is polar of C WRT the circle (U). CP meets (U) at P and again at Q. If P' is a reflection of P in $ CM \perp AB,$ $ P' \in (U)$ and CP' meets (U) again at Q'. PP'Q'Q is isosceles trapezod symmetrical WRT CM, the diagonals PQ', P'Q meet on CM and on the polar AB of C, $ M \equiv PQ' \cap P'Q \Longrightarrow$ $ \angle APM \equiv \angle APQ' = \angle BP'Q = \angle BPQ = 180^\circ - \angle BPC.$

We have $ \angle PBA=\angle PAC$. Let $ X$ be on $ PM$ such $ PAXB$ is parallogram. Then $ \angle XAB=\angle PAC,\angle XBA=\angle PBC$. Hence $ C,X$ are isogonal points wrt $ \triangle PAB$. Hence $ PC,PX$ is isogonal line $ \triangle PAB$. This imply $ \angle APM+\angle BPC=180$.

Lemma: Let $ABC$ be a triangle and $\Gamma $ its circumcircle. Let the tangent to $\Gamma $ at $B$ and $C$ intersect at $D$. Then $AD$ coincides with the $A$- symmedian of $\triangle ABC$. The proof can be found here. By angle chasing we can show that $AC$ and $BC$ are tangents to the circumcircle of $\triangle PAB$ at points $A$ and $B$ respectively. The tangents $AC$ and $BC$ intersect at $C$. So $PC$ coincides with the $C$- symmedian of $\triangle ABC$. Since $PM$ is the median of $\triangle ABC$, $\angle APM = \angle BPE$. or $\angle APM = 180^{\circ} - \angle BPC$. So $\angle APM + \angle BPC = 180^{\circ} $.

Official Solutions

A nice problem for symmedians We see that AC and AB are tangents to the circumcircle of $\triangle{APB} \implies CP $ is the symmedian. Then the rest easily follows from how symmedians are defined

Since $\triangle ABC$ is isosceles, we know that $\angle PAC=\angle PBA$. Consider $(APB)$. Since $\angle PAB=\angle PBC$ and $\angle PAC=\angle PBA$, lines $AC$ and $BC$ are tangent to $(APB)$. This implies that $PC$ lies on the $P$-symmedian. Let $CP$ intersect $AB$ at point $E$. By the definition of the symmedian, $\angle APM=\angle BPE$. Clearly $\angle BPE+\angle BPC=180^{\circ}$, so $\angle APM + \angle BPC = 180^{\circ}$.

Let $CP\cap AB=G$. Then the problem is equivalent to $\angle APM=\angle BPG$, aka. $CG$ is a symmedian in $\triangle APG$. Notice that $CB$ and $CA$ are the tangents to $(APB),$ so we are done.

Since $\angle CBP=\angle BAP$ and $\angle PAC=\angle PBA,$ $C$ is the intersection of the tangents to $(PAB)$ at $A$ and $B.$ Hence, $\overline{PC}$ is the $P$-symmedian point of triangle $PAB.$ Notice that $$\measuredangle BPC=-\measuredangle(\overline{CP},\overline{PB})=-\angle APM.$$$\square$

Nice problem! It is well known that $P$ is the $A$ humpty point, and so $M_1:=AP\cap BC$ is the midpoint of $BC$. Furthermore, we have that $\angle PCB=\angle PAC=\angle PBA$ and $\angle PAB=\angle PBC$. Hence, we get that $\angle APM+\angle BPC=\angle APM+\angle CPM_1+\angle M_1PB=\angle APM+\angle MPB+\angle BPM_1=180^{\circ},$ where this follows from similar triangles $\triangle PMB$ and $\triangle PM_1C$. EDIT: oops just realized I already solved this before. At least this is a slightly different/more motivated solution.

Let $(APB) \cap CP=L$ As well: $CA$ and $CB$ are tangent of $(APB)$ Then: $PQ$ is symedian of $(APB)$ $\implies \angle APM=\angle LPB \implies \angle APM + \angle BPC=180$

Let $CP$ meet circumcircle of $APB$ at S. Since $\angle PAB = \angle PBC$ then $CB$ is tangent to $APB$ and since $CB = CA$ then $SP$ is symmedian of $ASB$ so $\angle APM = \angle BPS$ and so $\angle APM + \angle BPC = \angle 180$.

Let $CP \cap AB = D$. The angle condition implies that $AC$ and $AB$ are tangent to the circumcircle of $APB$, then we know that $PD$ is the $P$-symmedian of $APB$, which implies that $\angle TPC = \angle APD = \angle MPB$ thus we have that $\angle BPC = \angle MPT$, which clearly gives us that $\angle APM + \angle BPC = 180$

Let $\overline{PM'}$ be a symmedian in triangle $BPC$, so it suffices to show that $A$ lies along the symmedian. But this is evident by the tangent intersection definition.

Let CP intersect AB at N, the circumradius of tri. ACP(center O) and tri. BCP(center O') be r, r' respectively. We have to prove that CP is P -symmedian in tri. APB and to show that we need to prove AN/BN=AP^2/BP^2 , We have [tri. APN]/[tri. BPN] = AN/BN = [tri. APC]/[tri. BPC]=(AP × sin<CAP)/(BP × sin<CBP) by sine rule, sin<CAP/sin<CBP=r'/r, by angle chasing we will get that tri. OO'P and tri. APB are similar so, r'/r = AP/BP then, we get that AN/BN=AP^2/BP^2 so, PN is P-symmedian in tri. APB. This implies <APM = <BPN; <BPN + <CPB=180°=> <APM + <BPC=180° .